Evaluate $\int\sec^3x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac12\sec x\tan x+\dfrac12\ln|\sec x+\tan x|+C\\$ (Choice B) B $\sec x\tan x+\ln|\sec x+\tan x|+C\\$ (Choice C) C $\dfrac12\sec x\tan x+\dfrac12\sec x+C\\$ (Choice D) D $\dfrac12\sec x\tan x+\sec x+C\\$
Explanation: In this problem we use integration by parts. Let $u=\sec x$ and $dv=\sec^2x\ dx$. Then $du=\sec x\tan x\,dx$ and $v=\tan x$. We now have $\int\sec^3x\,dx=\sec x\tan x-\int\sec x\tan^2 x\,dx$ Now use the identity $\tan^2x=\sec^2x-1$ to rewrite the integral on the right entirely in terms of $\sec x$. $\begin{aligned} &\phantom{=}\int\sec^3x\,dx \\\\ &=\sec x\tan x-\int\sec x(\sec^2x-1)\,dx \\\\ &=\sec x\tan x-\int\sec^3 x\,dx+\int\sec x\ dx \end{aligned}$ We now add $\int\sec^3x\,dx$ to both sides of the equation. $\begin{aligned} &\phantom{=}2\int\sec^3x\,dx \\\\ &=\sec x\tan x+\int\sec x\,dx \\\\ &=\sec x\tan x+\ln|\sec x+\tan x|+C \end{aligned}$ Finally, divide through by $2$. $\int\sec^3x\,dx=\dfrac12\sec x\tan x+\dfrac12\ln|\sec x+\tan x|+C$